4.57 Gears puts stress on....
Lugging the car at any time is a bad idea, particularly in a high gear. Whoever advised you to drive that way seriously misled you, S2KBreaker. Sorry to hear it.
Skip, yeah too bad you couldn't make it to the NorEaster. We had amazing weather and nice drives, and the autocross was a blast, as always.
Skip, yeah too bad you couldn't make it to the NorEaster. We had amazing weather and nice drives, and the autocross was a blast, as always.
Originally Posted by -KEY!S2K-,Aug 20 2006, 06:35 PM
I traded in a S2000 with a 4.77 final drive ratio. The Honda dealership never knew!
Originally Posted by raymo19,Aug 21 2006, 11:47 AM
I'm trying to wrap my mind around this but I can't. I'm no engineer by any means but I don't understand why the rear end would be stressed any more with 4.44s, 4.57s, or 4.77s. There's really nothing different going on past the diff than before.
The only difference is the engine and prop shaft are turning more RPMs to achieve the same ground speed.
Somebody care to elaborate?
The only difference is the engine and prop shaft are turning more RPMs to achieve the same ground speed.
Somebody care to elaborate?


That being said, I do believe fluid changes should not be tied to a finite mileage number. That is the gheyest thing I've ever heard(all these years!
) It should be based more on how the vehicle is driven. If after running higher rpm's more frequently, one should adjust his/her preventative maintenance schedule.
The only stress would be the higher rpm cruising speed, which would really be negligable if you follow proper maintainance. The 4.57 gears give the car a mechanical advantage, which in most cases would reduce stress not increase it. Swapping your gears isn't going to cause a #4 cylinder failure, or any type of engine failure for that matter. 500 rpm difference at 80mph on a car that has a 9000rpm redline is nothing, or even an 8k rpm redline on the 2004+. And the 2004's have a taller 6th gear which would help out.
The only time this would change is if your hammering the throttle all the time, dropping the clutch etc... In which case it wouldn't matter which gears you had.
Of course if its improperly installed your hosed!
The only time this would change is if your hammering the throttle all the time, dropping the clutch etc... In which case it wouldn't matter which gears you had.
Of course if its improperly installed your hosed!
Originally Posted by S2kracka,Aug 21 2006, 04:19 AM
There are no tips, just treat the car like you would any other S2k. The launches are actually easier on the car with gears.
thats good to hear
raymo19 Posted on Aug 21 2006, 05:47 PM
Gears, any gears, transmission, diff, are torque multipliers.
You get more torque with less revolutions.
So there is more torque going through your driveshafts then before.
Not when you are just diving along.
When accelerating.
A gear can be seen as leverage.
Force x length = torque (think of your torque wrench).
A gear is connected on to a shaft. Lets call that the driving gear.
That shaft gets torque delivered to it (by the engine or whatever you want).
The diameter of the driving gear determines the amount of force on the teeth of that driving gear.
Torque / length = force (half the gear diameter = length).
So now the driven gear has a force on its teeth.
The driven gear also has a diameter.
Force x length = torque.
Driving gear has a diameter of 20.
Driven gear has a diameter of 50.
Input torque is 100.
That gives you:
100 / 10
(half the diameter) = 10 (force on the teeth of the driving gear).
10 is also the force on the teeth of the driven gear.
10 x 25 = 250 output torque.
See? Thats more.
It comes at a price.
Revolutions.
Gears also have teeth.
How many is determined by the diameterof the gear and the shape and size of the teeth.
That driving gear with a diameter of 20 may have 30 teeth.
The driven gear with a diamterer of 50 may have 75 teeth.
That first shaft gets it torque at 5000 rpm.
So the driving gear spins at 5000 rpm.
The gear ratio is 30 : 75 equals 1 : 2.5
That makes the driven gear spin at 2000 rpm.
So we start with 5000 rpm and 100 torque.
And we get 2000 rpm and 250 torque.
There is a relation between diameter and amount of teeth.
The driving gear of 20 may also have 40 teeth.
But then the driven gear would (almost must) have 100.
Because the teeth design has to match to mesh correctly.
The calculations work out the same:
40 : 100 = 1 : 2.5
With hypoid gears in your diff it works the same.
The difference in ratio (4.10 - 4.44 - 4.57 - 4.77) and therefore diameter gives more torque at lower output revs.
When I say "difference in diameter" it is barely noticable with hypoid gears.
I think (I'm not completely sure) that difference is where the contact point is.
So the physical dimentions may look the same, the place where the force is transmitted is further away from the centre of the ring gear => more lenght => more torque.
Less revs because the ratio is greater.
(I hope I'm right, otherwise my whole explanation is complete BS)
That higher torque is delivered through the Torsen at the drive shafts, CV's and wheel & tire.
The nature of a hypoid gearset is that the forces that the driving gear puts on the ring gear are also axial, so it pushes the ring gear away.
Thats why you "need" stronger bolts and bearing caps.
Pffff....
I hope I've got it right.
I think I do..
I'm trying to wrap my mind around this but I can't. I'm no engineer by any means but I don't understand why the rear end would be stressed any more with 4.44s, 4.57s, or 4.77s. There's really nothing different going on past the diff than before.
You get more torque with less revolutions.
So there is more torque going through your driveshafts then before.
Not when you are just diving along.
When accelerating.
A gear can be seen as leverage.
Force x length = torque (think of your torque wrench).
A gear is connected on to a shaft. Lets call that the driving gear.
That shaft gets torque delivered to it (by the engine or whatever you want).
The diameter of the driving gear determines the amount of force on the teeth of that driving gear.
Torque / length = force (half the gear diameter = length).
So now the driven gear has a force on its teeth.
The driven gear also has a diameter.
Force x length = torque.
Driving gear has a diameter of 20.
Driven gear has a diameter of 50.
Input torque is 100.
That gives you:
100 / 10
(half the diameter) = 10 (force on the teeth of the driving gear).10 is also the force on the teeth of the driven gear.
10 x 25 = 250 output torque.
See? Thats more.
It comes at a price.
Revolutions.
Gears also have teeth.
How many is determined by the diameterof the gear and the shape and size of the teeth.
That driving gear with a diameter of 20 may have 30 teeth.
The driven gear with a diamterer of 50 may have 75 teeth.
That first shaft gets it torque at 5000 rpm.
So the driving gear spins at 5000 rpm.
The gear ratio is 30 : 75 equals 1 : 2.5
That makes the driven gear spin at 2000 rpm.
So we start with 5000 rpm and 100 torque.
And we get 2000 rpm and 250 torque.
There is a relation between diameter and amount of teeth.
The driving gear of 20 may also have 40 teeth.
But then the driven gear would (almost must) have 100.
Because the teeth design has to match to mesh correctly.
The calculations work out the same:
40 : 100 = 1 : 2.5
With hypoid gears in your diff it works the same.
The difference in ratio (4.10 - 4.44 - 4.57 - 4.77) and therefore diameter gives more torque at lower output revs.
When I say "difference in diameter" it is barely noticable with hypoid gears.
I think (I'm not completely sure) that difference is where the contact point is.
So the physical dimentions may look the same, the place where the force is transmitted is further away from the centre of the ring gear => more lenght => more torque.
Less revs because the ratio is greater.
(I hope I'm right, otherwise my whole explanation is complete BS)
That higher torque is delivered through the Torsen at the drive shafts, CV's and wheel & tire.
The nature of a hypoid gearset is that the forces that the driving gear puts on the ring gear are also axial, so it pushes the ring gear away.
Thats why you "need" stronger bolts and bearing caps.
Pffff....
I hope I've got it right.
I think I do..
Originally Posted by SpitfireS,Aug 21 2006, 03:48 PM
Gears, any gears, transmission, diff, are torque multipliers.
You get more torque with less revolutions.
You get more torque with less revolutions.
If you replace the stock 4.10 final drive with anything shorter (4.30, 4.44, etc.) when are you lowering RPM? It seems to me you're progressively pushing the stress to the front of the drivetrain and making the rear easier to turn.
I'll repeat my disclaimer about not being an engineer and add to it that I'm not a physicist either.
raymo19 Posted on Aug 22 2006, 12:27 AM
Note : numbers are for illustration only
And there is no such thing as friction loss.
Engine = 5000 rpm / 100 torque.
Input transmission = 5000 rpm / 100 torque.
Output transmission = 2000 rpm / 250 torque.
Input diff = 2000 rpm / 250 torque.
Output diff 4.10 = 487 rpm / 1025 torque.
Output diff 4.57 = 437 rpm / 1142 torque.
That torque goes through the output shafts, CV's, wheel & tire.
At the wheel you divide torque by half the diameter of the wheel & tire to get force.
Force where the tire touches the road.
Because action = reaction and your car can't put up the reaction to counter the force, you go forward.
(or backwards when you are in reverse
)
So when you step on it when your engine is at 5000 rpm you get more torque at the wheel with a 4.57 diff.
You are at a slower speed when you start stepping on it then a car with a 4.10 diff in the same gear at 5000 rpm.
When you are both at the same speed you (with 4.57) will be at higher rpm.
More rpm => more torque => 4.57 diff => even more torque
You win.
Until you have to shift because you run out of rpm's.
And then we start a whole new discussion.
But I'm not
Spits, I've read your post three times and still can't get past the second sentence.
Note : numbers are for illustration only
And there is no such thing as friction loss.
Engine = 5000 rpm / 100 torque.
Input transmission = 5000 rpm / 100 torque.
Output transmission = 2000 rpm / 250 torque.
Input diff = 2000 rpm / 250 torque.
Output diff 4.10 = 487 rpm / 1025 torque.
Output diff 4.57 = 437 rpm / 1142 torque.
That torque goes through the output shafts, CV's, wheel & tire.
At the wheel you divide torque by half the diameter of the wheel & tire to get force.
Force where the tire touches the road.
Because action = reaction and your car can't put up the reaction to counter the force, you go forward.
(or backwards when you are in reverse
)So when you step on it when your engine is at 5000 rpm you get more torque at the wheel with a 4.57 diff.
You are at a slower speed when you start stepping on it then a car with a 4.10 diff in the same gear at 5000 rpm.
When you are both at the same speed you (with 4.57) will be at higher rpm.
More rpm => more torque => 4.57 diff => even more torque
You win.
Until you have to shift because you run out of rpm's.
And then we start a whole new discussion.
But I'm not
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